Let $\mathbf x$ be a vector, and assume we have a distribution with mean $\mathbf \mu$ and covariance matrix $S$. The Mahalanobis distance with respect to the distribution is defined as
$$D(\mathbf x) = \sqrt{(\mathbf x – \mathbf \mu)^TS^{-1}(\mathbf x – \mathbf \mu)}.$$
Why? Well, $S$ is positive-definite, so we can decompose $\mathbf x – \mu$ into $\sum a_i\mathbf e_i$, where $\mathbf e_i$ are the orthonormal eigenvectors of $S$ (with corresponding eigenvalues $\lambda_i$). Then $S^{-1} a_i \mathbf e_i = \frac{a_i \mathbf e_i}{\lambda_i}$ and so
$$D(\mathbf x) = \sqrt{\sum \frac{a_i^2}{\lambda_i}}.$$
So the Mahalanobis distance takes into account how far out from the mean we are, where the meaning of “far out” depends on the direction you’ve traveled from the mean.